Math Word Problem?

A bag contains some $1, $5, $10 coins.
Its known that no. of $1 coins is 4 times the no. of $10 coins,
while the no. of $5 coins is less than that of $1 coins by 3.
If the total value of the coins is $121, find the no. of $10 coins.
x + 5y + 10z = 121
x = 4z
y = x - 3
Solution:
x = 16, y = 13, z = 4
The number of $10 coins is 4.

Let n be the number of $10 coins.
Then, number of $1 coins = 4n
and number of $5 coins = 4n - 3

Total volume of the coins in $:
10n + 1(4n) + 5(4n - 3) = 121
10n + 4n + 20n - 15 = 121
34n = 136
n = 4

Hence, number of $10 coins = 4

Where do they have a $10 coin?
x = number or $1 coins
x+10x/4+5(x-3) = 121
x = 16
Answer 4 $10 coins

Four $10 coins

Let 'a' be the number of $1 coins
Let 'b' be the number of $5 coins
Let 'c' be the number of $10 coins

It is known that the number of $1 coins is 4 times the number of $10 coins
a = 4c

It is known that the number of $5 coins is less than that of $1 coins by 3
b = a - 3

The total value of the coins is $121
1a + 5b + 10c = 121

Sub 'a' from your 1st equation and 'b' from your 2nd equation into your 3rd equation:
1a + 5b + 10c = 121
1(4c) + 5(a-3) + 10c = 121
4c + 5a - 15 + 10c = 121

Sub in 'a' from your 1st equation into your current equation:
4c + 5(4c) - 15 + 10c = 121
4c + 20c - 15 + 10c = 121
34c - 15 = 121
34c = 136
c = 4

Therefore the number of $10 coins is 4.

Extra:
a = 4c = 4*4 = 16
b = a - 3 = 16 - 3 = 13